how to shift trapped zeros to the bottom keeping leading zeros in given matrix fixed?
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Sushil Pokharel
on 21 Mar 2022
Commented: Sushil Pokharel
on 22 Mar 2022
Hi all, I have a problem while finding the probablity transition matrix. The code I have written to find the transition matrix calculate the transition probability matrices of each columns (in the given matrix below) but it excludes the transition between two states if zeros are trapped like
1
0
2 (shown below in bold) So, I wanted shift the trapped zeros to last.
I encountered trapped zeros in the matrix for instance, a matrix like
A = [0 0 0 1 0 2 3 1 2 3 0 0 0;
1 2 2 1 1 2 3 1 2 3 1 3 4;
0 1 0 1 0 0 0 1 2 1 1 2 2;
2 1 2 1 0 0 0 0 1 1 1 1 1;
1 0 0 1 1 1 1 1 2 2 2 1 1]
Now, I want to change this matrix into new matrix like
new_A =[0 0 0 1 0 2 3 1 2 3 0 0 0;
1 2 2 1 1 2 3 1 2 3 1 3 4;
2 1 2 1 1 1 1 1 2 1 1 2 2;
1 1 0 1 0 0 0 1 1 1 1 1 1;
0 0 0 1 0 0 0 0 2 2 2 1 1]
shifting all trapped zeros to the bottom of each column. Any help will be greatly appreciated.
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Accepted Answer
Matt J
on 21 Mar 2022
Edited: Matt J
on 21 Mar 2022
A=[ 0 0 0 1 0 2 3 1 2 3 0 0 0
1 2 2 1 1 2 3 1 2 3 1 3 4
0 1 0 1 0 0 0 1 2 1 1 2 2
2 1 2 1 0 0 0 0 1 1 1 1 1
1 0 0 1 1 1 1 1 2 2 2 1 1];
B=A>0;
C=cummax(B,1);
D=double(B);
D(C&A==0)=inf;
D=sort(D,1);
D=D~=0 & ~isinf(D);
Anew=zeros(size(A));
Anew(D)=A(A~=0)
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More Answers (1)
Arif Hoq
on 21 Mar 2022
if your matrix A almost same dimension( 5 X 13) everytime, then
A = [0 0 0 1 0 2 3 1 2 3 0 0 0;
1 2 2 1 1 2 3 1 2 3 1 3 4;
0 1 0 1 0 0 0 1 2 1 1 2 2;
2 1 2 1 0 0 0 0 1 1 1 1 1;
1 0 0 1 1 1 1 1 2 2 2 1 1];
B=A(:,1);
C=B(2:end);
C([2 end])=C([end 2]);
D=[B(1);C];
output=[D A(:,2:end)]
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